Single increment interval: [2kπ-π/2, 2kπ+π/4], k ∈ z;

Simple subtraction interval: [2kπ-3π/4, 2kπ-π/2, k∈z 。

Solution:

Let t=sinx+cosx=√2sin(x+π/4) and t be [-√ 2, √2].

Then sinxcosx =1/2 * (t 2-1),

∴y= 1/2*(t^2- 1)+t+2= 1/2*t^2+t+3/2

= 1/2(t+ 1)^2+ 1

∫t is in [-√2, √2], the opening is upward, and the symmetry axis t=- 1∈[-√2, √2],

There is a minimum value when t=- 1, and the minimum value = 1.

There is a maximum when t=√2, and the maximum value = (5+2 √ 2)/2.

∴y=sinxcosx+sinx+*cosx+2 range: [1, (5+2√2)/2]

①∫t∈[- 1, √2] single increase

∴- 1≤√2sin(x+π/4)≤√2

∴2kπ-π/2≤x≤2kπ+π/4,k∈z

②t∈[-√2,-1] single reduction

∴-√2≤√2sin(x+π/4)≤- 1

∴2kπ-3π/4≤x≤2kπ-π/2,k∈z

To sum up: required range: [1, (5+2√2)/2],

The growth interval of the function: [2kπ-π/2, 2kπ+π/4], k∈z 。

Simple decreasing interval of function: [2kπ-3π/4, 2kπ-π/2, k∈z 。